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Chapter 7: Problem 31

ssm Batman (mass \(=91 \mathrm{~kg}\) ) jumps straight down from a bridge into a boat (mass \(=510 \mathrm{~kg}\) ) in which a criminal is fleeing. The velocity of the boat is initially \(+11\) \(\mathrm{m} / \mathrm{s}\). What is the velocity of the boat after Batman lands in it?

Short Answer

Expert verified
The velocity of the boat after Batman lands is approximately \(+9.33 \, \text{m/s}\).

Step by step solution

01

Identify the System and Known Values

Batman jumps directly into the boat, so we need to consider conservation of momentum for the system consisting of Batman and the boat. The mass of Batman is given as \(m_1 = 91 \, \text{kg}\) and the mass of the boat is \(m_2 = 510 \, \text{kg}\). The initial velocity of the boat is \(v_{2i} = +11 \, \text{m/s}\). Batman's initial velocity is \(v_{1i} = 0 \, \text{m/s}\) since he is just dropping into the boat.
02

Apply the Principle of Conservation of Momentum

The principle of conservation of momentum states that the total momentum before Batman lands must equal the total momentum after he lands. The equation is:\[m_1 \times v_{1i} + m_2 \times v_{2i} = (m_1 + m_2) \times v_f\]where \(v_f\) is the final velocity of the boat after Batman lands.
03

Substitute the Known Values into the Equation

Substitute the known values into the momentum conservation equation:\[91 \, \text{kg} \times 0 \, \text{m/s} + 510 \, \text{kg} \times 11 \, \text{m/s} = (91 \, \text{kg} + 510 \, \text{kg}) \times v_f\]This simplifies to:\[0 + 5610 \, \text{kg} \cdot \text{m/s} = 601 \, \text{kg} \times v_f\]
04

Solve for the Final Velocity

Solve for \(v_f\) by dividing both sides of the equation by the total mass (601 kg):\[v_f = \frac{5610 \, \text{kg} \cdot \text{m/s}}{601 \, \text{kg}}\]\[v_f \approx 9.33 \, \text{m/s}\]
05

Interpret the Result

The final velocity of the boat, with Batman in it, is approximately \(+9.33 \, \text{m/s}\). This indicates that the boat slows down slightly after Batman lands in it, which makes sense because the total mass of the boat increased.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Momentum
Linear momentum is an essential concept in physics that involves understanding how objects move. It is defined as the product of an object's mass and velocity.
For any object, its linear momentum is calculated using the formula: \( p = m \times v \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.
  • If an object is in motion, it carries momentum that depends on how fast it is moving and how much it weighs.
  • The greater the mass or velocity of the object, the higher its momentum.
Momentum is conserved in closed systems, meaning that total momentum before an event must equal total momentum after the event. This principle is the foundation for solving collision-related problems.
Collision
A collision occurs when two or more objects come into contact and exert forces on each other. There are various types of collisions, like elastic and inelastic collisions.
In our exercise, Batman jumping into the boat represents an inelastic collision.
  • Inelastic collisions are characterized by the merging or sticking together of objects after collision.
  • Momentum is conserved, but kinetic energy is not necessarily conserved in inelastic collisions.
This means the total momentum of Batman and the boat before and after the collision remains constant. Understanding collisions helps us predict how objects will move and interact.
Velocity
Velocity is the speed of an object in a specified direction. It is different from speed as it includes direction; therefore, it is a vector quantity.
In our problem, the velocity is positive, showing the direction of the boat's motion.
  • Initial velocity, as given in this exercise, is the velocity of the criminal's boat before Batman lands (11 m/s).
  • Final velocity is what we calculate after the collision (approximately 9.33 m/s).
The velocity changes in the system because of the added mass after Batman joins the boat, resulting from the conservation of momentum.
Mass
Mass is a measure of how much matter is in an object. In this exercise, we deal with two masses: Batman's and the boat's.
Mass plays a crucial role in determining an object's momentum.
  • The mass of Batman is 91 kg, and for the boat, it is 510 kg.
  • When Batman joins the boat, their masses combine to form a single system mass, altering the boat's momentum and velocity.
This combination increases the overall mass in the system, resulting in a reduced velocity to maintain conserved momentum.

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Most popular questions from this chapter

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. (a) During the downward motion of the ball, are any of the following conserved: its momentum, its kinetic energy, its total mechanical energy? (b) During the collision with the block, are any of the following conserved: the horizontal component of the total momentum of the ball/block system, the total kinetic energy of the system? Provide reasons for your choices. Problem The masses of the ball and block are, respectively, \(1.60 \mathrm{~kg}\) and \(2.40 \mathrm{~kg},\) and the length of the wire is \(1.20 \mathrm{~m}\). Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

One average force \(\overline{\vec{F}}_{1}\) has a magnitude that is three times as large as that of another average force \(\overline{\mathrm{F}}_{2} .\) Both forces produce the same impulse. The average force \(\overline{\mathrm{F}}_{1}\) acts for a time interval of \(3.2 \mathrm{~ms}\). For what time interval does the average force \(\overline{\mathrm{F}}_{2}\) act?

At illustrates how to model a problem similar to this one. An automobile has a mass of \(2100 \mathrm{~kg}\) and a velocity of \(+17 \mathrm{~m} / \mathrm{s}\). It makes a rear-end collision with a stationary car whose mass is \(1900 \mathrm{~kg} .\) The cars lock bumpers and skid off together with the wheels locked. (a) What is the velocity of the two cars just after the collision? (b) Find the impulse (magnitude and direction) that acts on the skidding cars from just after the collision until they come to a halt. (c) If the coefficient of kinetic friction between the wheels of the cars and the pavement is \(\mu_{k}=0.68,\) determine how far the cars skid before coming to rest.

A car (mass \(=1100 \mathrm{~kg}\) ) is traveling at \(32 \mathrm{~m} / \mathrm{s}\) when it collides head-on with a sport utility vehicle (mass \(=2500 \mathrm{~kg}\) ) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?

A cannon of mass \(5.80 \times 10^{3} \mathrm{~kg}\) is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires an \(85.0-\mathrm{kg}\) shell horizontally with an initial velocity of \(+551 \mathrm{~m} / \mathrm{s}\). Suppose the cannon is then unbolted from the earth, and no external force hinders its recoil. What would be the velocity of a shell fired by this loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system.)
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